Math 307 - Spring 2025

Sunday, June 1, 2025

On Thursday and Friday we explored an amazing function \[ \forall\mkern 1mu x \in \mathbb{R} \qquad x \mapsto \frac{\sin (x)+\tanh (x)}{1 + \sin (x) \tanh (x)}. \]

In the Mathematica code below, I will consistently write this function as a pure function:
```
(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]) &
```
I find this preferable to assigning it a name, since it keeps us constantly aware of the function's spirit: Sin[#]& and Tanh[#]&.

Before preceding with the explorations of the given function, let us review what we know about its components

Plot[Sin[#] &[x], {x, -2 Pi, 2 Pi}, PlotStyle -> {Thickness[0.005]}, 
 Epilog -> {{PointSize[0.0125], RGBColor[0.7 {0, 0, 1}], 
    Point[{#, Sin[#]}] & /@ 
     Range[-4 Pi + Pi/2, 4 Pi, 2 Pi]}, {PointSize[0.0125], 
    RGBColor[0.7 {0, 1, 0}], 
    Point[{#, Sin[#]}] & /@ Range[-4 Pi + (3 Pi)/2, 4 Pi, 2 Pi]}}, 
 Ticks -> {Range[-4 Pi, 4 Pi, Pi/2], Automatic}]

Plot[Tanh[#] &[x], {x, -8, 8}, PlotStyle -> {Thickness[0.005]}, 
 Epilog -> {{Thickness[0.003], RGBColor[0.7 {0, 0, 1}], 
    Line[{{{-10, 1}, {10, 1}}}]}, {Thickness[0.003], 
    RGBColor[0.7 {0, 1, 0}], Line[{{{-10, -1}, {10, -1}}}]}}, 
 Ticks -> {Range[-10, 10, 1], Automatic}]

We see that Tanh[#]& is an odd sigmoid function with the horizontal asymptotes at $y=1$ and $y=-1$ and that Tanh[#]& approaches the asymptotes very rapidly. In fact \begin{align*} \tanh(20) &\approx 0.99999999999999999150329148941682204543855819119099\\ \tanh(50) &\approx 0.99999999999999999999999999999999999999999992559848 \end{align*}

A basic Plot[...] results in
```
 
Plot[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#])&[x], {x, -9, 9}, PlotStyle -> {Thickness[0.005]}]
```
The graph indicates that the function is odd: \[ \forall\mkern 1mu x \in \mathbb{R} \quad \frac{\sin (-x)+\tanh (-x)}{1 + \sin (-x) \tanh (-x)} = \frac{-\sin (x)-\tanh (x)}{1 + \sin (x) \tanh (x)} = - \frac{\sin (x)+\tanh (x)}{1 + \sin (x) \tanh (x)} \] Thus for all the features we can focus on the positive real numbers.
Natural questions arise: Is this function defined for all real numbers? What is the exact range of this function?
To understand this function we have to use our knowledge of its components Sin[#]& and Tanh[#]&.

We know that
```
 
Sin[#] & /@ Range[-10 Pi + Pi/2, 10 Pi, 2 Pi]
```
will result in all 1s. While
```
 
Sin[#] & /@ Range[-10 Pi + 3 Pi/2, 10 Pi, 2 Pi]
```
will result in all -1s.

Therefore it is interesting to evaluate our given function at those values. And we can do that "on paper". For every integer $k$ we have \[ \frac{\sin \bigl(\tfrac{\pi}{2}+2 k \pi\bigr)+\tanh(\tfrac{\pi}{2}+2 k \pi\bigr)}{1 + \sin(\tfrac{\pi}{2}+2 k \pi\bigr) \tanh(\tfrac{\pi}{2}+2 k \pi\bigr)} = \frac{1+\tanh(\tfrac{\pi}{2}+2 k \pi\bigr)}{1 + 1 \tanh(\tfrac{\pi}{2}+2 k \pi\bigr)} = \Large 1 \] and \[ \frac{\sin \bigl(\tfrac{3\pi}{2}+2 k \pi\bigr)+\tanh(\tfrac{3\pi}{2}+2 k \pi\bigr)}{1 + \sin(\tfrac{3\pi}{2}+2 k \pi\bigr) \tanh(\tfrac{3\pi}{2}+2 k \pi\bigr)} = \frac{-1+\tanh(\tfrac{\pi}{2}+2 k \pi\bigr)}{1 + (-1) \tanh(\tfrac{\pi}{2}+2 k \pi\bigr)} = - \Large 1 \]

There is a warning in the denominator of the last expression \[ 1 + (-1) \tanh(\tfrac{\pi}{2}+2 k \pi\bigr) = 1 - \tanh(\tfrac{\pi}{2}+2 k \pi\bigr). \] We have seen earlier that $\tanh(50)$ is very close to $1$. So when the positive integer $k$ is large, $\tanh(\tfrac{\pi}{2}+2 k \pi\bigr)$ is very close to $0$. It can be so close to $0$ that Mathematica can mistakenly take it to be $0$. We need to be aware of this in out calculations.

Let us implement what we calculated in the preceding item on a graph of the given function.

 
Plot[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]) &[x], {x, 0, 30}, 
 PlotStyle -> {Thickness[0.005]}, 
 Epilog -> {{PointSize[0.0125], RGBColor[0.7 {0, 0, 1}], 
    Point[{#, (Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#])}] & /@ 
     Range[-4 Pi + Pi/2, 40 Pi, 2 Pi]}, {PointSize[0.0125], 
    RGBColor[0.7 {0, 1, 0}], 
    Point[{#, 
        FullSimplify[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#])]}] & /@ 
     Range[-4 Pi + (3 Pi)/2, 40 Pi, 2 Pi]}}, 
 PlotRange -> {{0, 30}, {-1.1, 1.1}}, 
 Ticks -> {Range[-4 Pi, 40 Pi, Pi/2], Automatic}]

This graph clearly indicates that something suspicious is going on with this graph. What is happening near the $x$-values $\dfrac{3\pi}{2}+2 k \pi$ for nonnegative values of the integer $k$?

This needs additional explorations.

I decided that it will be a good idea to zoom in near the $x$-values \[ \dfrac{3\pi}{2}, \quad \dfrac{3\pi}{2}+2 \pi, \quad \dfrac{3\pi}{2}+4 \pi, \quad \dfrac{3\pi}{2}+6 \pi, \ldots \]

To automate this process, I introduced the variable nn for the number of $2\pi$ that I am adding and the variable dx for the range of the variable $x$ that I am allowing. So, I graph the given function on the interval

 
{x, 2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx}

I choose the following PlotRange->

 
PlotRange -> 
{{2 nn Pi + (3 Pi)/2 - dx,  2 nn Pi + (3 Pi)/2 + dx}, {-1.1, 1.1}}

In the command below I use Module[] to make the variables nn and dx local.

Module[{nn, dx}, nn = 1; dx = (3/2) 10^(-4); 
 Plot[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]) &[x] , {x, 
   2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx}, 
  PlotPoints -> 1500, PlotStyle -> {Thickness[0.005]}, 
  PlotRange -> {{2 nn Pi + (3 Pi)/2 - dx, 
     2 nn Pi + (3 Pi)/2 + dx}, {-1.1, 1.1}}, 
  Epilog -> {{Text[
      2 nn Pi + (3 Pi)/2, {2 nn Pi + (3 Pi)/2, 0.15}]}, {PointSize[
      0.01], RGBColor[0, 0.6, 0], 
     Point[{#, -1}] & /@ 
      Range[2 (nn - 1) Pi + (3 Pi)/2, 2 (nn + 1) Pi + (3 Pi)/2, 
       2 Pi]}}, Axes -> True, 
  Ticks -> {{#, N[#]} & /@ 
     Range[2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx, dx/4], 
    Automatic}, ImageSize -> 800]]

Module[{nn, dx}, nn = 2; dx = (3/2) 10^(-6); 
 Plot[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]) &[x] , 
  {x, 2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx}, 
  PlotPoints -> 1500, PlotStyle -> {Thickness[0.005]}, 
  PlotRange -> {{2 nn Pi + (3 Pi)/2 - dx, 
     2 nn Pi + (3 Pi)/2 + dx}, {-1.1, 1.1}}, 
  Epilog -> {{Text[
      2 nn Pi + (3 Pi)/2, {2 nn Pi + (3 Pi)/2, 0.15}]}, {PointSize[
      0.01], RGBColor[0, 0.6, 0], 
     Point[{#, -1}] & /@ 
      Range[2 (nn - 1) Pi + (3 Pi)/2, 2 (nn + 1) Pi + (3 Pi)/2, 
       2 Pi]}}, Axes -> True, 
  Ticks -> {{#, N[#]} & /@ 
     Range[2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx, dx/4], 
    Automatic}, ImageSize -> 800]]

Module[{nn, dx}, nn = 3; dx = (3/2) 10^(-8); 
 Plot[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]) &[x] , {x, 
   2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx}, 
  PlotPoints -> 1500, PlotStyle -> {Thickness[0.005]}, 
  PlotRange -> {{2 nn Pi + (3 Pi)/2 - dx, 
     2 nn Pi + (3 Pi)/2 + dx}, {-1.1, 1.1}}, 
  Epilog -> {{Text[
      2 nn Pi + (3 Pi)/2, {2 nn Pi + (3 Pi)/2, 0.15}]}, {PointSize[
      0.01], RGBColor[0, 0.6, 0], 
     Point[{#, -1}] & /@ 
      Range[2 (nn - 1) Pi + (3 Pi)/2, 2 (nn + 1) Pi + (3 Pi)/2, 
       2 Pi]}}, Axes -> True, 
  Ticks -> {{#, N[#]} & /@ 
     Range[2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx, dx/4], 
    Automatic}, ImageSize -> 800]]

But, unfortunately, this strategy does not work for nn = 3. The last graph above is not very informative. Even worse, when we zoom in to dx = (3/2) 10^(-9), Mathematica reports errors. I illustrate the errors just by numerical calculations below.

 
Module[{nn, dx}, nn = 3; dx = (3/2)*10^-(9); 
 N[ (Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#])] & /@ 
  Range[2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx, dx/4]]

Resulting in

But, by carefully considering our calculations above we understand the source for these errors. Mathematica thinks that it encounters zero in the denominator. It is possible that we overcome this problem if we increase the precision with which Mathematica does the calculations. We do that in the next item.

To overcome the calculational problem at the end of preceding item, we use SetPrecision[]
```
 
Module[{nn, dx}, nn = 3; dx = (3/2)*10^-(9); 
 N[SetPrecision[(Sin[#] + Tanh[#])/(1 + Sin[#] Tanh[#]), 100]] & /@ 
  Range[2 nn Pi + (3 Pi)/2 - dx, 2 nn Pi + (3 Pi)/2 + dx, dx/4]]
```
Resulting in

These numbers are not extreme. The extremely small numbers are in the denominator of the given function.

I was not able to use SetPrecision[] in Plot[]. A solution that I found is in the next item.
To be continued.

Wednesday, May 28, 2025

I am fascinated with envelopes, in particular those that lead to curves evolutes. In the next two items I show evolutes of the cycloid and the cardioid.
The images below show how the evolute of the cycloid emerges as the envelope of the normal lines drawn to the cycloid. What is unique about the cycloid is that its evolute is again a congruent cycloid shifted down by $2$ and to the right by $\pi$.

Click on the image below to see how the cycloid's evolute emerges.
The images below show how the evolute of the cardioid emerges as the envelope of the normal lines drawn to the cardioid. As one can see, the evolute of the cardioid is again a cardioid similar to the original cardioid. The transformation that maps the evolute to the original cardioid is given by the following matrix \[ \begin{bmatrix} -3 & 0 \\ 0 & 3 \end{bmatrix}. \] That is the evolute is flipped across the $y$-axis and then scaled by $3$.

Click on the image below to see how the cardioid's evolute emerges.

Tuesday, May 27, 2025 (updated)

In one of the problems on the next Assignment 2, I will ask you to explore the Cardioid and its generalizations.

The following table presents a derivation of the equation of a cardioid. For optimal viewing, use a wide screen.

Click through several helpful images.

The origin $O = (0,0)$ of the coordinate system is the gray point. Denote by $C$ the black point in the picture.
Both circles in the picture are unit circles, the gray one centered at $O,$ the black one centered at $C.$
The angle at the gray point marked by the blue arc between the green and the blue point measures $t$ radians. That is, the length of this blue arc is $t.$
There are two other angles in the picture which also measure $t$ radians: the angle marked by the blue arc between the blue and the red point and the angle marked by the black arc between the blue point and the purple point.
The green point is $(1,0),$ the blue point is $(\cos t, \sin t),$ the black point $C$ is $(2\cos t, 2\sin t),$ the purple point is $(2\cos t-1, 2\sin t).$
The goal is to determine the coordinates of the red point.
Denote by $\mathbf{a}$ the unit vector whose tail is the black point and whose head is the red point. Based on the previous items we have \begin{align*} \mathbf{a} & = \bigl(\cos( \pi+2 t ) \bigr) \mathbf{i} + \bigl(\sin( \pi+2 t ) \bigr) \mathbf{j} \\ & = -\bigl(\cos(2 t ) \bigr) \mathbf{i} - \bigl(\sin(2 t ) \bigr) \mathbf{j}. \end{align*}
Denote by $\mathbf{c}$ the position vector of the black point. Then the position vector of the red point is \[ \mathbf{c} + \mathbf{a} = \bigl( 2 \cos t - \cos(2 t ) \bigr) \mathbf{i} + \bigl(2 \sin t - \sin(2 t ) \bigr) \mathbf{j}. \]
Thus the parametric equations for this cardioid are \begin{align*} x & = 2 \cos t - \cos(2 t ) \\ y & = 2 \sin t - \sin(2 t ) \end{align*}

In the Cardioid part of Assignment 2 you need to reproduce two pictures from Wikipidia's Cardioid page I made an animation that unifies these two pictures. You should be able to produce something like the animation below . If not, you can present one animation and one picture.
Place the cursor over the image to start the animation.
Once you have figured out how to construct a cardioid, you should be able to construct Generalized Cardioids. In this part of Assignment 2 you are asked to produce some of the generalized cardioids as shown in the animations below . These animated gif files are large. On a slow internet connection it takes a while for them to load.
Below is the cardioid generated by a wheel with radius 1/2 rolling on a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 1/3 rolling on a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 2 rolling on a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 3/2 rolling on a circle with radius 1.
Place the cursor over the image to start the animation.
For completeness, I post four more generalized cardioids. The animations below are large. On a slow internet connection it takes a while for them to load.
Below is the cardioid generated by a wheel with radius 1/2 rolling inside of a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 1/3 rolling inside of a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 2 rolling "inside" of a circle with radius 1.
Place the cursor over the image to start the animation.
Below is the cardioid generated by a wheel with radius 3/2 rolling "inside" of a circle with radius 1.
Place the cursor over the image to start the animation.

Saturday, May 24, 2025

I am fascinated how the circle involute appear as an envelope of a family of circles. Below I created 9 pictures with a different number of circles in the family whose envelope is the circle involute.

Recall that this family consists of the circles centered at $\bigl( \cos(t) , \sin(t) \bigr)$ and with the radius $t$, where $t$ is any positive integer. The code for the Manipulate[] command that produces images like displayed below.
```
 
Manipulate[Graphics[
  {(* enveloped circles  *)
   {RGBColor[0.65 {1, 1, 1}], Thickness[0.001], 
    Circle[{Cos[#], Sin[#]}, #] & /@ Range[Pi/nn, 4 Pi, Pi/nn]} 
   },
  PlotRange -> {{-11.6, 13.6}, {-12.6, 12.6}}, 
  AspectRatio -> Automatic, Background -> White, 
  PlotLabel -> TableForm[{nn*4, "circles"}, TableDirections -> Row],
  Frame -> False, ImageSize -> 800
  ], {{nn, 16}, {1, 2, 4, 6, 8, 16, 32, 48, 64}, 
  ControlPlacement -> Top, ControlType -> Setter}]
```
Click on the image below to see the circle involute appear.
We deduced the parametric vector equation for the circle involute to be, in the traditional calculus notation: \[ x(t) = \cos(t) + t \sin(t), \ \ y(t) = \sin(t) - t \cos(t) \quad \text{with} \quad t \gt 0, \] or in the traditional vector notation \[ \begin{bmatrix} \cos(t) + t \sin(t) \\[5pt] \sin(t) - t \cos(t) \end{bmatrix} = \begin{bmatrix} \cos(t) \\[5pt] \sin(t) \end{bmatrix} + t \begin{bmatrix} \sin(t) \\[5pt] - \cos(t) \end{bmatrix}, \quad \text{with} \quad t \gt 0. \] To plot the circle involute in Mathematica, we use the following command:
```
ParametricPlot[
 {Cos[t] + t Sin[t], -t Cos[t] + Sin[t]}, {t, 0, 4 Pi}, 
 PlotStyle -> {{RGBColor[0.5 {0, 1, 1}], Thickness[0.004]}},
 PlotRange -> {{-13.1, 10.1}, {-13.1, 10.1}}, Axes -> False, 
 AspectRatio -> Automatic, Background -> White,
 Frame -> False, ImageSize -> 800
 ]
```
In this item, I want point out how the unit circle can be reveled from the equation, or the graph of its involute.

The first step is to identify the tangent line at each point of the involute. The derivative of the vector equation of the circle involute will give us the direction vector of the tangent line at each point: \[ \frac{\mathrm{d}}{\mathrm{d} t} \begin{bmatrix} \cos(t) + t \sin(t) \\[5pt] \sin(t) - t \cos(t) \end{bmatrix} = \begin{bmatrix} t \cos(t) \\[5pt] t \sin(t) \end{bmatrix}, \quad \text{with} \quad t \gt 0. \] In the picture below we use this vector to construct the red tangent line to the involute at a specific point.

The second step is to construct the normal line at each point of the involute. To get the vector normal to the tangent vector we apply the rotation matrix by $\pi/2$: \[ \begin{bmatrix} 0 & -1 \\[5pt] 1 & 0 \end{bmatrix} \begin{bmatrix} t \cos(t) \\[5pt] t \sin(t) \end{bmatrix} = \begin{bmatrix} - t \sin(t) \\[5pt] t \cos(t) \end{bmatrix}, \quad \text{with} \quad t \gt 0. \] In the picture below we use this normal vector to construct the gray normal line to the involute at a specific point.

The code that we used for the picture is
```
ParametricPlot[
 {Cos[t] + t Sin[t], -t Cos[t] + Sin[t]}, {t, 0, 4 Pi}, 
 PlotStyle -> {{RGBColor[0.5 {0, 1, 1}], Thickness[0.004]}},
 Epilog -> {{ Thickness[0.002], RGBColor[{1, 0, 0}],
    Line[{{Cos[#] + # Sin[#], -# Cos[#] + Sin[#]} - 
         20 {Cos[#], Sin[#]}, {Cos[#] + # Sin[#], -# Cos[#] + 
           Sin[#]} + 20 {Cos[#], Sin[#]}}] &[10]},
   {Thickness[0.002], RGBColor[0.75 {1, 1, 1}], 
    Line[{{Cos[#] + # Sin[#], -# Cos[#] + 
          Sin[#]}, {Cos[#] + # Sin[#], -# Cos[#] + Sin[#]} + 
         1.2 {-# Sin[#], # Cos[#]}}] &[10]},
   {PointSize[0.007], 
    Point[{Cos[#] + # Sin[#], -# Cos[#] + Sin[#]}] &[10]}},
 PlotRange -> {{-13.1, 10.1}, {-13.1, 10.1}}, Axes -> False, 
 AspectRatio -> Automatic, Background -> White,
 Frame -> False, ImageSize -> 800
 ]
```

In this item we take advantage that all code in the preceding item is written as pure functions. So, now it is straightforward to draw many normal lines to the circle involute. To avoid clutter, we did not draw the points and the red tangent lines.

ParametricPlot[
 {Cos[t] + t Sin[t], -t Cos[t] + Sin[t]}, {t, 0, 4 Pi}, 
 PlotStyle -> {{RGBColor[0.5 {0, 1, 1}], Thickness[0.002]}},
 Epilog -> {{Thickness[0.002], RGBColor[0.75 {1, 1, 1}], 
    Line[{{Cos[#] + # Sin[#], -# Cos[#] + 
          Sin[#]}, {Cos[#] + # Sin[#], -# Cos[#] + Sin[#]} + 
         1.2 {-# Sin[#], # Cos[#]}}] & /@ 
     Range[Pi/Sqrt[5], 4 Pi, (4 Pi - Pi/Sqrt[5])/40]}},
 PlotRange -> {{-13.1, 10.1}, {-13.1, 10.1}}, Axes -> False, 
 AspectRatio -> Automatic, Background -> White,
 Frame -> False, ImageSize -> 800
 ]

In conclusion, we recovered the unit circle by drawing many normal lines to the circle involute.

The above fact, expressed using the concept of an envelope: The unit circle is the envelope of the family of all normal lines to the circle involute.

In general, we can start from any curve, draw the family of its normal lines, and look for the envelope of that family. In most cases, the envelope exists, and it is called the evolute of the original curve.

The pictures above illustrate the fact that the unit circle is the evolute of the circle involute.

Thursday, May 22, 2025

Yesterday, I posted about highly useful concept in Mathematica: the pure function. In my code, pay attention to the occurrences of pure functions and how I use them to create many lines or many points or many circles.

Pure Function. In earlier exercises, we often assigned a name to a function for further use. However, this approach is not always practical. Mathematica allows us to use functions as if they were named without actually naming them. For example, the square function is not explicitly named in Mathematica, but we can calculate the square of 7 by writing (#^2)&[7]. In this case, we use (#^2)& as a reference to the square function. Instead of explicitly naming the argument, we use a placeholder #, and we conclude the definition with &.

Notice that, instead of explicitly naming the argument (independent variable), in a pure function we use # which is called a slot in Mathematica. It acts as a placeholder for the argument that will be passed to the function. The symbol & is called function operator in Mathematica. It indicates the end of a pure function definition. For example, in the pure function (#^2)& the expression (#^2) defines the function, and ampersand & marks the end of the function.

The full potential of a pure function is realized when combined with the Map[] command. For instance, Map[(#^2)&,{a,b,c}] yields {a^2,b^2,c^2}. Thus, Map[] applies the specified function to each member of the given list. An alternative shortcut form for Map[] is /@. Using this shortcut, the previous example becomes (#^2)&/@{a,b,c}. This approach is extremely beneficial when dealing with functions given with several expressions and large lists containing hundreds of entries.
On Assignment 2 I will ask you to explore two famous curves: the cycloid and the cardioid. First we discuss the cycloid.
Wikipedia's definition of a Cycoid is: A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. One could demonstrate a cycloid in a dark room by attaching a small light to a rolling wheel and taking a long exposure photography of the motion. This is what American photographer Berenice Abbott did to create her photograph Cycloid:

You can read more about Berenice Abbott and her art at MoCP and Mount Holyoke College
Cycloid. Derivation of the equation. In this item I will illustrate the derivation of the parametric equations for the cycloid which you need to implement in the animations in Mathematica.
In the following derivation the vectors $\mathbf{i}$ and $\mathbf{j}$ are the unit coordinate vectors $\displaystyle \left[\begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\displaystyle \left[\begin{array}{c} 0 \\ 1 \end{array} \right]$, respectively. In Mathematica notation the vector $\mathbf{i}$ is the pair {1,0} and the vector $\mathbf{j}$ is the pair {0,1}.

Let $t\in [0,\pi].$ The position vector of the gray center of the circle in the above pictures is \[ t \, \mathbf{i} + \mathbf{j}. \] If the unit circle was positioned at the origin, then the position vector of the red point on the unit circle would be \[ \bigl(\cos(-t -\pi/2) \bigr) \, \mathbf{i} + \bigl(\sin(-t -\pi/2) \bigr) \mathbf{j} = - (\sin t ) \, \mathbf{i} - (\cos t ) \mathbf{j}. \] Consequently, the position vector of the red point on the unit circle in the above pictures is \[ \bigl( t - \sin t \bigr) \, \mathbf{i} + \bigl(1 - \cos t \bigr) \mathbf{j}. \] In Mathematica notation the vector $\mathbf{i}$ is the pair {1,0} and the vector $\mathbf{j}$ is the pair {0,1}. Thus, the parametric equations of the cycloid in Mathematica notation is
```
 
(t - Sin[t]) {1,0} + (1 - Cos[t]) {0,1} 
```
Or, simplified,
```
 
{t - Sin[t] , 1 - Cos[t] } 
```
The following animation might help you to understand the above derivation.

Place the cursor over the image to start the animation.
In one of the problems on Assignment 2 I will ask you to reproduce the following three animations.
Place the cursor over the image to start the animation.

Place the cursor over the image to start the animation.

Place the cursor over the image to start the animation.

Wednesday, May 21, 2025

What I presented on Tuesday is an introduction to the construction of the cycloid and the cardioid.

The cycloid and the cardioid are just two famous planar curves. There are many, many more. Visit this excellent site with over 350 planar curves: mathcurve.com — Planar Curves
On Tuesday we studied the unit circle and its tangents.
Although, I did not introduce the concept of pure function in class, I will use it here and I will introduce this concept tomorrow in class.
A convenient tool in Mathematica constructions is the concept of pure function.

Pure Function. In the preceding exercises we often named them for further use. However, this is not always practical. Mathematica allows us to use the functions as if they were named without naming them. For example the square function is not named in Mathematica, but we can write (#^2)&[7] to get the square of 7. In some sense, we use (#^2)& as a name of the square function. Instead of naming the argument we are giving it a "place holder" # and we finish the definition with &.

The full power of pure function is realized in combination with the command Map[]. For example Map[(#^2)&,{a,b,c}] gives {a^2,b^2,c^2}. Thus, Map[] applies the specified function to the given list. Alternative shortcut form for Map[] is /@. In the previous simple example it would be (#^2)&/@{a,b,c}. This is extremely useful when we have a complicated function and a large list with hundreds of entries.

Next we play with the unit circle and its tangent lines. In the spirit of the above introduced concept of pure function, I define a tangent line to the unit circle as a pure function:

Line[{
     {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
         {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}
          }]&

To understand this definition you need to think of points as vectors. The following vector,

{Cos[#], Sin[#]}

is the position vector of a point on the unit circle. Instead of a variable name we put # . The vector

{-Sin[#], Cos[#]}

is orthogonal to the vector specified above. Always think of points as vectors. The following pair of points

{{Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},  
             {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}

are on the tangent line to the unit circle. This tangent line touches the circle at the point

{Cos[#], Sin[#]}

This is illustrated in the picture below:

The complete code for the above picture is

 
Graphics[
 {
  {RGBColor[0, 0.5, 0.5], Thickness[0.005], Circle[{0, 0}, 1]},
 {
  Thickness[0.0015], RGBColor[0.7, 0.7, 0.7], 
   Line[{
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, 
            {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}] &[4 Pi/5]}
  }, 
  {
  PlotRange -> {{-5, 5}, {-5, 5}}, Frame -> True, 
  PlotRangeClipping -> True, Background -> Automatic, 
  ImageSize -> 800
  }
  ]

In the above code, notice how we applied pure function to produce a line:

 
   Line[{
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, 
            {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}] &[4 Pi/5]

Since we used the pure function to create the above image, it is easy to Map[] the pure function to a range of values to get a more interesting picture. There are two ways to implement Map[]. One is

Map[
Line[{
  {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
    {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}]&, Range[0,2 Pi,(2 Pi)/5]
]

The other way is to use the abbreviation /@

Line[{
  {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
    {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}
        }]&/@Range[0,2 Pi,(2 Pi)/5]

Below is the complete code for the picture of a unit circle with five tangent lines. Notice how we made just a small change to the preceding code.

 
Graphics[
 {
  {RGBColor[0, 0.5, 0.5], Thickness[0.005], Circle[{0, 0}, 1]},
  {Thickness[0.0015], RGBColor[0.7, 0.7, 0.7], 
   Line[
   {
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, {Cos[#], Sin[#]} 
              - 10 {-Sin[#], Cos[#]}
              }]&/@ Range[(2 Pi)/5, 2 Pi, (2 Pi)/5]}
  },
 {PlotRange -> {{-5, 5}, {-5, 5}}, Frame -> True, 
  PlotRangeClipping -> True, Background -> Automatic, 
  ImageSize -> 800}]

Changing the step in the Range[] function, from (2 Pi)/5 to Pi/128, we get an interesting image below

Or, changing the interval in the Range[] function from the original interval 0,2 Pi to 0,(3/2) Pi we get an interesting image

Or, setting the interval in the Range[] function to be 0,Pi we get an interesting image

In this item I present some variations on the preceding item. Extending tangents from the touching point in one direction we get Or, shortening the length of the touching line segments And some artistic images inspired by this theme
If you think that I was just playing in the preceding items, just as an option to consider, I did not. The preceding considerations naturally extend towards the construction of the involute of the unit circle.

Think of the circle below as an idealized roll of scotch tape. How is tape being unwrapped in a geometrically ideal way?

To answer the above question, we use the idea from the preceding work. Say we unwrap a piece of the tape of length $\pi/3.$ This is the corresponding geometrically ideal picture (click on the image to start an animation of the unwrapping of one full layer of tape)

The involute of the unit circle is the curve formed by the trace of the blue point in the preceding animation. In Mathematica we need to apply Line[] to the list of all points which are colored dark blue in the previous animation. For the animation hover cursor over the image.
Mathematica's command which created the complete involute of the unit circle in the preceding image is
```
{RGBColor[0, 0, 0.7], Thickness[0.003],
 Line[({Cos[#], Sin[#]} - # {-Sin[#], Cos[#]})&/@Range[0,3.28 Pi,0.02]]}
```
The corresponding traditional vector formula for the circle involute is: \[ \bigl( \cos(t) + t \sin(t), \sin(t) - t \cos(t) \bigr) \quad \text{with} \quad t \geq 0. \]
It is quite fascinating that the involute of the unit circle is related to the family of circles which are presented below as teal circles in the next image:
Mathematica's command for the teal circles is
```
{RGBColor[0, 0.6, 0.6], Thickness[0.001],
         Circle[{Cos[#], Sin[#]},#]& /@ Range[0,(5-1/6) Pi,Pi/48]}
```
In the next animation I present only the circles. You can see how the circles on their own recreate the involute. In fact, the involute is the envelope of this family of circles. This means that the involute is tangent to each of the circles in this family.

In the last scene of the preceding animation there are 233 circles. Below I show the image of 77 circles and even with fewer circles one can see the envelope phenomena of the involute.

Below I show the image of 38 circles and even with fewer circles one can see the envelope phenomena of the involute.
Another variation on the involute and the family of circles whose envelope is the involute is to present the circles and the involute as a 3d-image. We lift each circle at the level $z$ which equals the radius of the circle. In this way we form a surface in the image below:
```
 Show[ParametricPlot3D[{Cos[z] + z Cos[t], Sin[z] + z Sin[t], z},
  {t,0,2 Pi}, {z,0,4 Pi}, Mesh -> {0, 4*24}, MeshStyle -> {None, RGBColor[0,0.6,0.6]},
  PlotPoints -> {100, 100}, PlotStyle -> {Opacity[0.2]}],
 ParametricPlot3D[{Cos[z] + z Sin[z], Sin[z] - z Cos[z], z},
 {z,0,4 Pi}, PlotPoints -> {200}, PlotStyle -> {RGBColor[0, 0, 0.7]}],
 BoxRatios -> {1, 1, 1}, Boxed -> False, Axes -> False, ImageSize -> 800]
```
The following animation illustrates the process of creating the surface described above:

The bird's eye view of the 3d-surface that we created is exactly one of the images that we created earlier.

Sunday, May 18, 2025

It is important in life to orient oneself in the world of COLORS. To help with that, I wrote a webpage: Color Cube. And all illustrations on the Color Cube page are created in Mathematica.
Being able to navigate the world of C O L O R S is important. To help with this, I created a page: Color Cube. All illustrations on the Color Cube page were generated using Mathematica.
One of the remarkable applications of trigonometric functions Cos[t] and Sin[x] is that they describe small oscillations of a stretched string. In the animation below you can see the function \[ \cos(t) \sin(x) \qquad \text{with} \qquad x \in [0,\pi] \quad \text{and} \quad t \in [0,2\pi] \] As $t$ takes values from $0$ to $2\pi$, the function $\sin(x)$ is scaled with a different number $\cos(t)$ between $1$ and $-1$. In the animation below I take 241 values of $t$ between, and including, $0$ to $2\pi$, create 241 images and use Mathematica to export those 241 images as an animated gif file.

Hover over the image to start the animation.

This is the Mathematica code used to create the above animation and export it as an animated gif file.

(* create a list of 240 images of an oscillating sine function, call \
this list SinxCost *) 
SinxCost = 
  Table[Plot[Evaluate[Sin[x] Cos[t]], {x, 0, Pi}, 
    PlotStyle -> {{Thickness[0.01], RGBColor[1, 0, 0]}},
    Epilog -> {
      {PointSize[0.012], White, Point[#] & /@ {{0, 0}, {Pi, 0}}},
      {Text["t =", {Pi/2 - 0.1, 1.26}, 
        BaseStyle -> {FontWeight -> "Normal", 
          FontColor -> RGBColor[1, 1, 1]}], 
       Text[NumberForm[N[t], {3, 2}], {Pi/2, 1.26}, 
        BaseStyle -> {FontWeight -> "Normal", 
          FontColor -> RGBColor[1, 1, 1]}]}
      },
    PlotRange -> {{-0.1, Pi + 0.1}, {-1.4, 1.4}}, AspectRatio -> 1/5, 
    Frame -> True, FrameTicks -> {{{}, {}}, {Range[0, Pi, Pi/4], {}}},
     Axes -> False, ImageSize -> 800, Background -> Black], {t, 0, 
    2 Pi, 2 Pi/240.}]; 

(* decide which duration you want for each frame, here 0.1 and create \
a table of these durations, call it dds *)

dds = 0.1 & /@ Range[Length[SinxCost]]; 

(* Set the directory where you want the movie to be exported. *) 

SetDirectory["C:\\Dropbox\\Work\\myweb\\Courses\\Math_pages\\Math_307\
\\Pictures"]

(* Export the list of 240 images as an animated gif file *) 

Export["SinxCostAni.gif", SinxCost, 
 "AnimationRepetitions" -> Infinity, "ImageSize" -> 800,  "DisplayDurations" -> dds]

(* Export the first image as a gif file *) 

Export["SinxCostS1.gif", SinxCost[[1]], "ImageSize" -> 800]

Since we worked so much with the cosine and sine function, I started playing with the graph of the sine function. I moved it around, placed it climbing along the $y$-axis. Then, I decided to animate each of the graphs and this turns into art.

Hover over the image to start the animation.

Hover over the image to start the animation.

Hover over the image to start the animation.

Saturday, May 17, 2025

In the PNG picture below I explain the geometry that is behind the definition of NgonCos[] and NgonSin[] which I defined yesterday in class. In the PNG picture below I used the function Mod[t,1]. It is just the function t - Floor[t].
On Friday, we introduced functions inspired by the trigonometric functions cosine and sine, but designed to draw regular ngons, for example equilateral triangle, square, pentagon, hexagon, heptagon, octagon, nonagon, decagon, hendecagon, dodecagon, and so on. The notebook that I created on Friday you can find in our shared Dropbox directory \307_Files\2025\ under the name 20250516_n-gons.nb.
The most important feature of the functions Cos[t] and Sin[t] is that they provide a parametrization of the unit circle. An animation illustrating this feature was posted on Friday, May 9.
The principle presented in the animation posted on Friday, May 9 is the general principle of parametric equations of curves in the plane. Given two functions (one green and one blue) of the same independent variable t, taken as the coordinates of a moving point, then, as the independent variable t changes, the point does trace a curve in the plane.

The formulas derived in the first item in this post, implemented in Mathematica are as follows:

Clear[NgonCos, NgonSin, t, nn];
NgonCos[t_, nn_] 
:= (1 - (t - Floor[t])) Cos[(2/nn) \[Pi] Floor[t]] 
        + (t - Floor[t]) Cos[(2/nn) \[Pi] (1 + Floor[t])];
NgonSin[t_, nn_] 
:= (1 - (t - Floor[t])) Sin[(2/nn) \[Pi] Floor[t]] 
         + (t - Floor[t]) Sin[(2/nn) \[Pi] (1 + Floor[t])];

The final product of this work are the graphs of the first nine regular $n$-gons inscribed in the unit circel: the equilateral triangle, the square, and the pentagon, the hexagon, the heptagon, the octagon, the nonagon, the decagon, the hendecagon, the dodecagon, see the Regular Polygon webpage at Wolfram MathWorld.

Friday, May 9, 2025

Assignment 1 has been posted in our shared Dropbox directory Dropbox\307_Files\Assignments.
To appreciate the Beauty of Trigonometry, it is essential to understand the key feature of the trigonometric functions Cos[t] and Sin[t]:
- The trigonometric functions Cos[t] and Sin[t] provide a parametrization of the unit circle. But that is just the beginning—most importantly, a deep understanding of the sphere and torus is impossible without Cos[t] and Sin[t].
The animation below is my attempt to illustrate how the parametrization of the unit circle works.
The main components of the animation are as follows:
- The black circle is the unit circle.
- The values of t $\in [0,2\pi]$ are represented by orange lengths. In each scene there are three orange lengths: the horizontal length, the vertical length and the arc length on the unit circle.
- The values of Cos[t] are green horizontal line segments.
- The values of Sin[t] are blue vertical line segments.

Hover over the image to start the animation.

The principle above is the general principle of parametric equations of curves in the plane. Given two functions (one green and one blue) of the same independent variable t, if these functions define the coordinates of a moving point, then as the independent variable t varies, the point traces a curve in the plane.

Monday, May 5, 2025

The Mathematica part of the class will start on Monday, May 5, 2025 in BH 215 at 10:00 am.
The information sheet
We will use
To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica. Watching the movies is essential for being able to organize your homework notebooks well! I will not discuss the basics of notebook structuring in class.
In the study of the computational software system Wolfram Mathematica, we will focus on symbolic and numerical calculations, visualization, and programming in Mathematica. But Mathematica is much more than that. To read more about Mathematica see An Elementary Introduction to the Wolfram Language.
As we begin the Mathematica portion of this class, a natural question arises: Why learn Mathematica? This is not an easy question to answer, but below are some key aspects that make Mathematica worth your attention.
- Mathematica is a powerful computational tool that can handle a wide range of mathematical problems, from basic algebra to advanced calculus and beyond.
- Unlike many other mathematical software systems, Mathematica excels in symbolic computation, allowing you to manipulate mathematical expressions and equations symbolically, closely resembling abstract mathematical thinking and structures.
- Mathematica offers powerful visualization tools, allowing you to create compelling graphics that enhance the understanding of mathematical concepts and effectively illustrate abstract mathematical results, fostering a varied and engaging mathematical learning experience.
- Mathematica is a high-level programming language that offers an extensive array of predefined functions, designed to simplify and streamline tasks that are commonly encountered in mathematical and technical computing. My impression is that this makes it more accessible and user-friendly compared to other mathematical software, enabling users to achieve complex objectives with greater ease.
- A lot of help is available on-line, for example Mathematica Stack Exchange, Wolfram Community and ChatGPT is quite good at offering help with Mathematica questions.
- Mathematica makes solving mathematical problems more efficient. How? For problems involving symbolic manipulation, we can perform symbolic manipulations in Mathematica exactly. For problems in which symbolic manipulations are inefficient, numerical approximations can be performed. Additionally, powerful graphical visualizations help guide the problem-solving process, offering deeper insights.
I have enjoyed the interaction between my mind and computers since I first started using them in the early 1990s. Recently, I came across a powerful metaphor for this relationship, expressed by Steve Jobs. In an old interview, he illustrated the usefulness of computers with a compelling analogy: "For me, a computer has always been a bicycle for the mind." That perfectly describes Mathematica and other modern tools like ChatGPT. Isn't that wonderful?
See the Bicycle for the Mind video. Or, a longer version.
I am convinced that your mathematical learning experience will be enriched if you engage in mathematical explorations using technology.
I recently encountered a fascinating geometric fact online: Imagine two squares where a vertex of the larger square is positioned at the center of the smaller square. Regardless of how these two squares are positioned, the area of their intersection remains constant. My description might not fully capture the beauty of this fact — indeed, an animation is worth a thousand words. How am I to make an animation? I must use technology. Here is an animation created in Mathematica.

Hover over the image to start the animation.
I should have left it as an exercise for you to figure out why what I claim in the preceding item is true. But, the solution supported by a Mathematica animation might indicate why using technology can be powerful tool in communicating Mathematics.
Hover over the image to start the animation.

Spring 2025
MATH 307: Mathematical computing
with Mathematica

Branko Ćurgus

Spring 2025 MATH 307: Mathematical computing with Mathematica

Branko Ćurgus

Spring 2025
MATH 307: Mathematical computing
with Mathematica